Question

please help and i reward a medal

For what value of x is sin x = cos 19°, where 0°< x < 90°?

A.38°
B.71°
C.26°
D.19°

Answer 1

Take note of the following property:
\[\sin \alpha=\cos 90-\alpha \] or the other way round.
So\[x=90-19=71\]
Answer, B.

Answer 2

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